Let → a = α ^ i + ^ j - ^ k and → b = 2 ^ i + ^ j - α ^ k , α > 0 ....

Given, \overset{\to}{a} = \alpha\overset{^}{i} + \overset{^}{j} - \overset{^}{k} and \overset{\to}{b} = 2\overset{^}{i} + \overset{^}{j} - \alpha\overset{^}{k}, \alpha > 0. If the projection of $\overset{\to}{a} \times \overset{\to}{b}$ on the vector - \overset{^}{i} + 2\overset{^}{j} - 2\overset{^}{k} is $30$, then $\alpha$ is equal to

Now \overset{\to}{a} \times \overset{\to}{b} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ \alpha & 1 & - 1 \\ 2 & 1 & - \alpha\end{cases}\right| = \left(1 - \alpha\right)\overset{^}{i} + \left({\alpha}^{2} - 2\right)\overset{^}{j} + \left(\alpha - 2\right)\overset{^}{k}

Projection of $\overset{\to}{a} \times \overset{\to}{b}$ on - \overset{^}{i} + 2\overset{^}{j} - 2\overset{^}{k}

= \left|\dfrac{\left(\overset{\to}{a} \times \overset{\to}{b}\right) \cdot \left(- \overset{^}{i} + 2\overset{^}{j} - 2\overset{^}{k}\right)}{\sqrt{{1}^{2} + {2}^{2} + {2}^{2}}}\right|

= \dfrac{\left(\left(1 - \alpha\right)\overset{^}{i} + \left({\alpha}^{2} - 2\right)\overset{^}{j} + \left(\alpha - 2\right)\overset{^}{k}\right) . \left(- \overset{^}{i} + 2\overset{^}{j} - 2\overset{^}{k}\right)}{3}

$= \dfrac{- 1 + \alpha + 2{\alpha}^{2} - 4 - 2\alpha + 4}{3}$

$= \dfrac{2{\alpha}^{2} - \alpha - 1}{3}$

Now given length of projection is $30$

So $\dfrac{2{\alpha}^{2} - \alpha - 1}{3} = 30$

$⇒ 2{\alpha}^{2} - \alpha - 91 = 0$

$⇒ \alpha = 7, - \dfrac{13}{2}$